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znimon

  • 2 years ago

Solve the non-linear inequality. Express the solution in interval notation. x^4 is greater than x^2

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  1. znimon
    • 2 years ago
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    \[x^4>x^2\]

  2. znimon
    • 2 years ago
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    I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x

  3. sriramkumar
    • 2 years ago
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    \[x ^{4}>x ^{2}\] \[x ^{4}-x ^{2}>0\] \[x ^{2}(x ^{2}-1)>0\] \[x ^{2}>0 or x ^{2}>1\] x>0 or x<-1 or x>1 then x<-1 and x>0 will be the interval...

  4. klimenkov
    • 2 years ago
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    \(x^2(x^2-1)>0\) divide it by \(x^2\): \((x-1)(x+1)>0\) \(x\in(-\infty;-1)\bigcup(1;\infty)\)

  5. znimon
    • 2 years ago
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    why is it greater than 0

  6. znimon
    • 2 years ago
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    oh subtract the x^2?

  7. sriramkumar
    • 2 years ago
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    yep//....

  8. sriramkumar
    • 2 years ago
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    @znimon understood the process?

  9. znimon
    • 2 years ago
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    almost, what about after x^2(x^2-1) is greater than 0

  10. klimenkov
    • 2 years ago
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    \(x^2\) is always bigger than 0. It equals 0 when \(x=0\). But it doesn't satisfy the inequality.

  11. sriramkumar
    • 2 years ago
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    took x^2 common in both the terms....

  12. znimon
    • 2 years ago
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    yep got that

  13. sriramkumar
    • 2 years ago
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    nops.. while splitting the inequality x has to be >0

  14. znimon
    • 2 years ago
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    will you break it down after the x^2(x^2-1) is greater than 0

  15. sriramkumar
    • 2 years ago
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    yep.... both the split terms are compared to the inequalities...

  16. znimon
    • 2 years ago
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    I understand how it works so I need to find all the solutions for x that satisfy the inequality

  17. znimon
    • 2 years ago
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    thanks now I'll try some on my own. Will you recommend a problem that has the same properties?

  18. znimon
    • 2 years ago
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    @sriramkumar

  19. sriramkumar
    • 2 years ago
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    try this one... \[x ^{8}-x ^{2}<0\] remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1

  20. znimon
    • 2 years ago
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    thanks

  21. sriramkumar
    • 2 years ago
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    @znimon you're welcome :)))))))))))))))))

  22. oldrin.bataku
    • 2 years ago
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    \[\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 non-zero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ |x|\ge\sqrt1\]

  23. oldrin.bataku
    • 2 years ago
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    is either 0 or non-zero **

  24. znimon
    • 2 years ago
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    solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?

  25. znimon
    • 2 years ago
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    for x^3-x^2 is less than 0

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